Cylinder on moving wedge A cylinder of mass m is rolling down, without slip- ping, on a wedge of mass M. The wedge is freely sliding on a horizontal table without friction and the axis of the cylinder remains horizontal during the motion. Gravity acts vertically. 1. Using Newton's laws and the horizontal Cartesian coordinates of the axis of the cylinder and the wedge, x and X, find the acceleration of the cylinder and the wedge, X and X. 2. Using Largangian techniques and the relative position of the cylinder's axis on the wedges, find s and X. Expressing s in terms of x compare to your Newtonian solution. 1. Newtonian method Forces on cylinder: F gravity, Fwedge, Fric Forces on wedge: -Fwedge,-Ftriction where F=F cos(a) friction Equations of motion for center of masses F+F+Fw=mF -F-Fw=MR Equation of motion for rotation Fƒ cos(a) = I cos(a)ē R Where R is the radius of the cylinder. M m Constraint: the cylinder is rolling without slipping, so dx = dX + Rcos (a) de, that is x=x+R cos(a)ē Summing the first two equations of motion shows that F₁-mr-MR and projecting to the x direction -mx=MX X α shows that the horizontal center of mass is conserved - due to lack of external horizontal forces. Com- bining the second equation of motion (projected in the x direction) with the rotational one and the constraint, after eliminating x using the first equation of motion, we obtain gm sin(a) cos(a) k² (m+M)+M

Cylinder on moving wedge A cylinder of mass m is rolling down, without slip- ping, on a wedge of mass M. The wedge is freely sliding on a horizontal table without friction and the axis of the cylinder remains horizontal during the motion. Gravity acts vertically. 1. Using Newton's laws and the horizontal Cartesian coordinates of the axis of the cylinder and the wedge, x and X, find the acceleration of the cylinder and the wedge, X and X. 2. Using Largangian techniques and the relative position of the cylinder's axis on the wedges, find s and X. Expressing s in terms of x compare to your Newtonian solution. 1. Newtonian method Forces on cylinder: F gravity, Fwedge, Fric Forces on wedge: -Fwedge,-Ftriction where F=F cos(a) friction Equations of motion for center of masses F+F+Fw=mF -F-Fw=MR Equation of motion for rotation Fƒ cos(a) = I cos(a)ē R Where R is the radius of the cylinder. M m Constraint: the cylinder is rolling without slipping, so dx = dX + Rcos (a) de, that is x=x+R cos(a)ē Summing the first two equations of motion shows that F₁-mr-MR and projecting to the x direction -mx=MX X α shows that the horizontal center of mass is conserved - due to lack of external horizontal forces. Com- bining the second equation of motion (projected in the x direction) with the rotational one and the constraint, after eliminating x using the first equation of motion, we obtain gm sin(a) cos(a) k² (m+M)+M

Similar questions

All parts! A long thin rod of length L and mass M is connected to a hoop of mass M and radius L/2 to form a spoked wheel as shown in the figure below. The wheel is mounted to a fixed axle (frictionless) and can spin about an axis through the center of the wheel and perpendicular to plane of the wheel. Several turns of a light rope are wrapped around the wheel and a mass M is attached to the rope. Use this information to answer the next 3 questions. 1. What is the moment of inertia of the wheel about the rotation axis? (ML^2)/2 (ML^2/6) (ML^2/4) (ML^2)/12 (ML^2)/3 (3ML^2)/20 2. The mass is released from rest and falls downward. What is the tension in the rope as the mass falls? T = (Mg)/4 T = (Mg)/2 T = (2Mg)/3 T = (2Mg)/5 T = (4Mg)/7 T = (3Mg)/8 3. What is the angular acceleration of the wheel as the mass falls? a = (5g)/(4L) a = (6g)/(5L) a = g/L a = (6g)/(7L) a = (3g)/(2L) a = (2g)/(3L)
Please help me
A vertical rectangular object of uniform mass M = 7 kg has sides a = 0, 26 m and b = 0,58 m. It is pivoted from point O as shown in the figure below. If a constant force of F = 109 N along the diagonal is applied as shown below, determine the angular acceleration of the rectangle in SI units at the instant shown. Express your answer using one decimal place. Please use the convention: Clockwise (+), Counterclockwise (-) when expressing your answer. Hint 1: The moment of inertia of a uniform disk about an axis of rotation passing through the center of mass (CM) is ICM = 1½M(a²+b²). Hint 2: Don't forget the force of gravity!!! Take g = 9.80 m/s². a CM 0 b F
One end of a thin, uniform rod is connected to a frictionless hinge as shown in the figure below (Figure 1). The rod has a length of 0.8 mm and a mass of 2 kgkg. It is held up in the horizontal position (θ=90∘�=90∘) and then released. Calculate the angular velocity of the rod at θ=0∘�=0∘.
As shown in the figure below, we have a square one meter on a side that is free to rotate about an axis perpendicular to the plane of the square, a distance a from one side and a distance b from the other side. a = b = b (i) Two forces, and F2, are applied to diagonally opposite corners, and act along the sides of the square, first as shown in case (i) and then as shown in case (ii) of the drawing. In each case the net torque produced by the forces is zero. If the magnitude of F2 is 4 times that of F₁, find the distances a and b that locate the axis. It should be noted that a and b are not drawn to scale. m m b a
Help me please
What is the rotational kinetic energy for 1 kilogram wheel that is rolling along a horizontal surface at a speed of 6 meters/second?